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Thermodynamics - Isentropic Compression

  1. Nov 27, 2013 #1
    Hello all,

    Its review time again for another Thermodynamics midterm. As such, I have a practice exam to try for optional extra review work. I've come across a problem that i'm somewhat stumped on. I've tried the problem, but I feel as though I've made too many assumptions in trying to solve the problem. Here's what I got

    1. The problem statement, all variables and given/known data

    Air, an ideal gas, with temperature-dependent heat capacities, is to be compressed from 100 kPa to 1500 kPa. The air enters the compressor at a temperature of 300K, and the compressor has an isentropic efficiency of 85%.

    How much power is required to operate the compressor adiabatically using an inlet air flow rate of 2.0 m3/s? (Calculate Power in kW)

    2. Relevant equations

    p1 = 100 kPa
    T1 = 300 K

    p2 = 1500 kPa
    T2 = ?

    ηisentropic=.85

    min = 2.0m3/s

    3. The attempt at a solution

    To start the equation off, I set up the relation p2/p1 = pr2/pr1
    The problem gave us values for both p1 and p2.

    I looked up the value of temperature-dependent pr1 which turned out to be:
    pr1 = 1.3860

    While I was looking at the air tables, I also took the enthalpy value at 300K, which turned out to be:
    h1 = 300.19

    I plugged pr1 into the pressure/pressure-reduced relationship, found pr2 to come out to 20.79, roughly the 20.64 found at 640K on the same air table. From this, I pulled the secong enthalpy value:
    h2s = 649.22

    From here, I try to solve for h2 I set up the equation for isentropic efficiency:
    η = (h1 - h2)/(h1 - h2s)

    Solving for h2, I get:
    h2 = 596.8655

    Finally, I solve for power.

    The equation I use is:
    [itex]\dot{Q}[/itex] - [itex]\dot{W}[/itex] + (mi(hi)-me(he)
    **I cancelled out the kinetic and potential energy terms of the original equation**

    Since the process is adiabatic, I eliminate the [itex]\dot{Q}[/itex].

    Here's where I think I may have gone wrong

    I then assume that the process is running at steady state, thus mi = me = 2.0m3/s

    Plugging 2 in for m and the rest of the values in for hi, he, I find that [itex]\dot{W}[/itex] = -593.351kW

    ...this CAN'T be right. I'm quite possibly doing this whole problem wrong. Thus, I turn to the almighty physicsforums.com for help. I know this problem and my solution might be long-winded, so I appreciate all patience with this problem in advance. I also DEFINITELY appreciate any and all help/advice sent my way.

    Thank you for your time
    -Lou
     
  2. jcsd
  3. Nov 27, 2013 #2
    This problem is much easier than you think. The first thing to do is to calculate the mass flow rate of air, from its volumetric flow rate at the inlet and its inlet density, which you can determine from the inlet temperature and pressure.

    When they say that the isentropic efficiency is 85% in this context, what they mean is that the work done if the compressor were operated isentropicly divided by the actual amount of work done is 0.85 (for the same starting and ending pressures). So you need to find the isentropic work done (per unit mass). You seem to be working from air tables, so you need to look up the properties at the initial conditions, and you need to look up the properties at the final pressure for the same entropy as in the initial state. This will tell you the initial and final specific enthalpy for the isentropic operation of the compressor. You can use these to determine the amount of work for isentropic operation.
     
  4. Apr 4, 2014 #3
    Bumping with a slight variation on this problem.

    In a refrigerator I'm trying to calculate the change in specific enthalpy during compression (h2-h1).
    h2=f(h1, h2_is) where h2_is is of course the h2 of isentropic compression. From there I have no issue finding the real h2. The problem, however, is in finding h2_is.

    h2_is is equal to the enthalpy at p2 and s2=s1 (which is easily found by referencing the h1 and p1) in the overheated gas tables for my refrigerant, but since I am trying to calculate based on variable unput, I must find a formula for h2_is instead of a manual reference.

    For the saturated vapor tables this has been easy. Because one input is enough to find all outputs through the use of trendline formulas. For overheated gas, 2 independent variables are required to find any third, which creates a third dimension of reference.

    I have, aside from point 2 in the cycle, all temperatures, pressures, enthalpies, etc, so I should have all variables I need to find h2_is through the condition that s1=s2, but the formulas for (change in) specific entropy have been highly confusing at best. If there is a set s1 for a given p1 and h1, then surely there must be a way to calculate h2 out of s1 and p2.

    Thank you for your time,
    -A desperate student trying to graduate.
     
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